What is the (sin^3(t))^' ?

The (sin^3(t))^' is 3sin^2(t)cos(t)

What is the first (sin^3(t))^' ?

The first (sin^3(t))^' is 3sin^2(t)cos(t)

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(sin^{3} (t))^{^{'}}

3 sin^{2} (t) cos (t)

Solution steps

(sin^{3} (t))^{'}

Apply the chain rule:

3 (sin (t))^{2} (sin (t))^{'}

= 3 (sin (t))^{2} (sin (t))^{'}

(sin (t))^{'} = cos (t)

= 3 (sin (t))^{2} cos (t)

Simplify

= 3 sin^{2} (t) cos (t)

y^{^{''}} = - 4 y + e^{x}

\int \frac{x ^{2} - 3 x - 8}{x ^{2} - 2 x + 1} d x

x y - 6 y^{2} = 4, (14, 2)

\int x^{n - 1} a + b x^{n} d x

\frac{d}{d x} (x (x + 4)^{2})